Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $22.8$ years; the standard deviation is $2.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than $27.2$ years.
Explanation: $22.8$ $20.6$ $25$ $18.4$ $27.2$ $16.2$ $29.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $22.8$ years. We know the standard deviation is $2.2$ years, so one standard deviation below the mean is $20.6$ years and one standard deviation above the mean is $25$ years. Two standard deviations below the mean is $18.4$ years and two standard deviations above the mean is $27.2$ years. Three standard deviations below the mean is $16.2$ years and three standard deviations above the mean is $29.4$ years. We are interested in the probability of a zebra living less than $27.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $18.4$ years and the other half $({2.5\%})$ will live longer than $27.2$ years. The probability of a particular zebra living less than $27.2$ years is ${95\%} + {2.5\%}$, or $97.5\%$.